Chemistry Paper 1 Flashcards
(54 cards)
After 1 minute the mean mass = 0.60 After 2 minutes the mean mass = 1.20 After 4 minutes the mean mass = 2.40 After 5 minutes the mean mass = 3.06 Determine the mean mass of copper produced after 3 minutes
1.80 mg
This question is about electrolysis. A student investigates the mass of copper produced during electrolysis of copper chloride solution. Which gas is produced at the positive electrode (anode)?
Chlorine gas
This question is about electrolysis. A student investigates the mass of copper produced during electrolysis of copper chloride solution. Copper is produced at the negative electrode (cathode). What does this tell you about the reactivity of copper?
Copper is less reactive than hydrogen
Experiment 1 = 3.02 mg Experiment 2 = X Experiment 3 = 3.01 mg Mean of all three experiments = 3.06 mg Calculate the mass X of copper produced in experiment 2 after 5 minutes
- 01 + 3.02 + x / 3 = 3.06
- 06 x 3 = 9.18
- 18 - 3.01 = 6.17
- 17 - 3.02 = 3.15
- 15 mg
The copper chloride solution used in the investigation contained 300 grams per dm^3 of solid CuCl2 dissolved in 1dm^3 of water. The student used 50cm^3 of copper chloride solution in each experiment. Calculate the mass of solid copper chloride used in each experiment
50 / 1000 = 0.05
0.05 x 300 = 15
15 g
State one similarity and one difference in the electronic structure of sodium and chlorine
Similarity - Both have three levels. Both have incomplete outer shells.
Difference - Sodium has one outer electron shell whereas chlorine has seven. The number of electrons are both different.
Sodium atoms react with chlorine atoms to produce sodium chloride (NaCl). Describe what happens when a sodium atom reacts with a chlorine atom. Write about electron transfer in your answer.
Sodium atom loses one electron from its outer shell whereas a chlorine atom gains an electron on its outer shell. Oxidation is loss of electrons but gain of oxygen and reduction is gain of electrons but loss of oxygen.
A student plans a method to prepare pure crystals of copper sulphate. The student’s method is: 1. Add one spatula of calcium carbonate to dilute hydrochloric acid in a beaker. 2. When the fizzing stops, heat the solution with a bunsen burner until all the liquid is gone. The method contains several errors and does not produce copper sulphate crystals. Explain the improvements the student should make to the method so that pure crystals of copper sulphate are produced.
- The dilute hydrochloric acid should be warmed before calcium carbonate is added into a beaker. Use a pipette to add calcium carbonate to the acid solution to increase levels of accuracy.
- Keep adding calcium carbonate to the acid solution until it no longer reacts.
- Filter the excess calcium carbonate to leave a solution of salt.
- Gently warm the salt solution over a bunsen burner so that the water evaporates and crystals of salt are formed.
Hazards/risks:
- corrosive acid can cause damage to eyes so eye protection must be worn
- hot equipment can cause burns so care must be taken when the salt solution is warmed.
What is the total number of electrons in a fluorine molecule (F2)?
18 electrons
Aluminium reacts with bromine to produce aluminium bromide. Complete the balanced chemical equation for this reaction.
2Al + 3Br2 –> 2AlBr3
Explain why diamond has a high melting point
Diamond has a high melting point because it is made up of a strong, rigid covalent lattice. Each carbon atom forms four strong covalent bonds with other carbon atoms. All strong covalent bonds mean that it is a very hard substance with both high melting and boiling points. There are no charged particles so it does not conduct electricity. The bonds forming diamond must be broken for the substance to melt or boil.
The relative formula mass pf a group 2 metal carbonate is 197
Relative atomic masses: C = 12 O = 16
Calculate the relative atomic masses of the group 2 metal in the metal carbonate
3 x 16 = 48
48 + 12 = 60
197 - 60 = 137
137
A - colourless solution, no solid - pH 14
B - colourless solution, no solid - pH 3
C - colourless solution, solid remains - pH 9
D - colourless liquid, solid remains - pH 7
Identify the solids A, B, C and D. Explain your answers.
A - sodium oxide - because it has highest pH
B - phosphorus oxide - because it has lowest pH
C - magnesium oxide - because it has 2nd highest pH
D - silicon dioxide - because it is neutral
10cm^3 = 3
1000cm^3 = X
Calculate the value of X
dilution by a factor of 100
(pH=) 5
Iron reacts with dilute hydrochloric acid to produce iron chloride solution and one other product. Name the other product.
Hydrogen
Iron reacts with dilute hydrochloric acid to produce iron chloride solution and one other product. Suggest why any unreacted iron can be separated from the mixture.
Due to filtration.
Mganesium reacts with iron chloride solution. 3Mg + 2FeCl3 –> 2Fe + 3MgCl2. 0.120g of magnesium reacts with excess iron chloride solution. Relative atomic masses: Mg = 24 Fe = 56. Calculate the mass of iron produced in mg
0.12 / 24 = 0.005
0.005 x 2/3 = 0.00333
0.00333 x 56 = 0.1866 g
187 mg
Explain which species is reduced in the reaction between magnesium and iron chloride. Your answer should include the half equation for reduction.
Fe3+ + 3e(-) –>
This is due to the fact reduction is gain of electrons but loss of oxygen.
Figure one shows what happens when calcium, copper, magnesium and zinc are added to hydrochloric acid. What is the order of decreasing reactivity of these four metals?
Ca Mg Zn Cn
Figure one shows what happens when calcium, copper, magnesium and zinc are added to hydrochloric acid. Name two variables that must be kept constant.
- mass of metal / element
- surface area of metal /element
- concentration of acid
- volume of acid
- temperature of acid
Figure one shows what happens when calcium, copper, magnesium and zinc are added to hydrochloric acid. What is the independent variable in this reaction?
type of metal / element
Predict the reactivity of beryllium compared with magnesium.
Give a reason for your answer. Use the periodic table.
Beryllium is less reactive.
• greater attraction between nucleus and outer electrons
• more energy is needed to remove electrons
• loss of electrons is more difficult
• outer electrons closer to nucleus
• less shielding
A solution of hydrochloric acid contains 3.2 g of hydrogen chloride in 50 cm3. Calculate the concentration of hydrogen chloride in g per dm3
50 / 1000 = 0.05
3.2 / 0.05 = 64
64 g per dm3
Give the state symbol for ammonium nitrate solution.
(aq) - aqueous